Beam Deflection Formulas for Common Load Cases
Beam deflection is the transverse displacement of a structural member under load — and for most real designs it matters as much as stress. A mezzanine floor joist, a crane runway girder or a machine-frame crossmember must not sag enough to crack a ceiling finish, jam a bearing track or cause dynamic problems, regardless of whether the bending stress is within the allowable limit. Deflection limits (typically L/250 to L/500 of the span in EN 1993-1-1) are independent serviceability checks that run alongside the strength check, and failing either one sends you back to resize the section.
The closed-form Euler-Bernoulli expressions collected here give exact analytical deflections for the six load cases most commonly encountered in structural and mechanical engineering practice — simply supported and cantilever beams under central point loads or uniformly distributed loads, plus the fixed-fixed (clamped) configuration. These are the same formulas implemented in the MechanixCalc beam deflection calculator, which applies them together with the EN 1993-1-1 serviceability and strength framework, the Goodman fatigue criterion and a natural-frequency check, and produces a fully traceable branded PDF engineering report.
The governing equation and what flexural rigidity EI means
All closed-form beam deflection formulas come from integrating the Euler-Bernoulli differential equation EI·d²y/dx² = M(x) twice along the beam span with the boundary conditions that define the support type. The product EI — flexural rigidity — is the single most important quantity: E is the material's Young's modulus (how stiff the material is) and I is the second moment of area of the cross-section about the bending axis (how efficiently the material is distributed away from the neutral axis). Doubling the span increases deflection by 2³ = 8× for a point load; doubling EI halves it. Choosing a deeper section rather than a heavier one is almost always the correct trade-off, because I scales with depth cubed.
The second moment of area for standard cross-sections follows directly from geometry: for a solid circular bar of diameter d, I = π·d⁴/64; for a solid rectangle of width b and depth h, I = b·h³/12; for a hollow tube, subtract the inner section's I from the outer. I-sections are efficient in bending precisely because material is concentrated in the flanges, far from the neutral axis, where it contributes most to I.
I = π · d⁴ / 64where I = second moment of area (mm⁴); d = outer diameter (mm)
I = b · h³ / 12where I = second moment of area (mm⁴); b = section width (mm); h = section depth (mm) — depth is the dimension in the direction of bending
Deflection formulas for the six standard load cases
The table below gives the maximum deflection for each of the six load cases supported by the calculator. All formulas produce deflection in the same units as L (typically mm) when F is in N, w is in N/mm, E is in N/mm² and I is in mm⁴. The location of the maximum deflection is shown alongside each formula.
Simply supported beam, central point load: the maximum deflection occurs at midspan and is δ_max = F·L³/(48·E·I). This is the most-used case in structural checking. Simply supported beam, uniformly distributed load (UDL): the maximum deflection at midspan is δ_max = 5·w·L⁴/(384·E·I). The factor 5/384 ≈ 0.01302 is commonly memorised. Cantilever beam, end point load: the tip deflects by δ_max = F·L³/(3·E·I) — sixteen times more than the simply supported case for the same span, load and section, because the effective span is doubled and the boundary condition halves the restraint. Cantilever, UDL: the tip deflection is δ_max = w·L⁴/(8·E·I). Fixed-fixed beam, central point load: end fixity reduces the midspan deflection to δ_max = F·L³/(192·E·I) — four times stiffer than simply supported — but requires that the supports can develop the end moments M_A = M_B = F·L/8.
δ_max = F · L³ / (48 · E · I)where δ_max = maximum deflection (mm); F = point load (N); L = span (mm); E = Young's modulus (N/mm²); I = second moment of area (mm⁴)
δ_max = F · L³ / (3 · E · I)where δ_max = maximum tip deflection (mm); F = tip load (N); L = cantilever length (mm); E = Young's modulus (N/mm²); I = second moment of area (mm⁴)
δ_max = 5 · w · L⁴ / (384 · E · I)where δ_max = maximum deflection (mm); w = load intensity (N/mm); L = span (mm); E = Young's modulus (N/mm²); I = second moment of area (mm⁴)
δ_max = F · L³ / (192 · E · I)where δ_max = maximum deflection (mm); F = central point load (N); L = span (mm); E = Young's modulus (N/mm²); I = second moment of area (mm⁴); end moments M_A = M_B = F·L/8 must be resisted by the supports
Bending stress and the yield safety factor
Once the peak bending moment M_max is known from the load case, the maximum bending stress follows from the flexure formula σ = M·c/I, where c is the distance from the neutral axis to the outermost fibre of the cross-section (half the depth for a symmetric section). The bending stress must not exceed the yield strength Sy of the material; in practice a safety factor SF = Sy/σ_max of at least 1.5–2 is expected for most structural applications, and the governing code (EN 1993-1-1 for steel) sets the partial factor to apply.
For the simply supported central-load case, the peak bending moment at midspan is M_max = F·L/4. For a cantilever with an end load it is M_max = F·L at the root. For a UDL on a simply supported beam, M_max = w·L²/8 at midspan. These moment values are inputs to the stress check — the deflection check and the stress check are both required and are independent of each other.
σ = M · c / Iwhere σ = bending stress (MPa); M = bending moment (N·mm); c = distance from neutral axis to extreme fibre (mm); I = second moment of area (mm⁴)
SF = Sy / σ_maxwhere SF = yield safety factor (dimensionless); Sy = yield strength of the material (MPa); σ_max = maximum bending stress (MPa)
How to choose and apply the right formula
The two things to get right before applying any formula are the support condition and the load type. A 'simply supported' beam is one that is free to rotate at both ends but cannot translate vertically; this is a close model for a beam resting on two knife-edge supports, standard shelf bearings or pinned-end connections. A 'fixed' or 'clamped' end cannot rotate and cannot translate — this requires a moment-resisting connection such as a welded end plate or embedded foundation. In practice, most real connections are somewhere in between, so assuming simply supported is often the more conservative (larger deflection) assumption unless the fixity is deliberate and verified.
For spans carrying both a point load and a uniform distributed load simultaneously, the deflections and bending moments from the two cases may be added directly (superposition), because Euler-Bernoulli beam theory is linear. The MechanixCalc combined-loading panel applies superposition automatically for SS and cantilever beams. For two-span continuous beams the calculator applies the 3-moment (Clapeyron) equation, which accounts for the bending moment at the interior support coupling the two spans together.
Worked example
A simply supported steel beam of span L = 2000 mm carries a central point load F = 5 kN. The cross-section is a solid circular bar of diameter d = 80 mm (E = 210 GPa, yield strength Sy = 250 MPa). Find the maximum deflection, peak bending moment, peak bending stress and yield safety factor.
Given
- Span L2000 mm
- Central point load F5000 N (5 kN)
- Bar diameter d80 mm
- Young's modulus E210 000 N/mm² (210 GPa, steel)
- Yield strength Sy250 MPa
Result
- Maximum deflection δ_max1.97 mm
- Peak bending moment M_max2.50 kN·m
- Peak bending stress σ_max49.7 MPa
- Yield safety factor SF5.03 (pass)
- Compute the second moment of area: I = π·d⁴/64 = π × 80⁴/64 = π × 40 960 000/64 = π × 640 000 = 2 010 619 mm⁴ ≈ 2.011 × 10⁶ mm⁴. The extreme-fibre distance is c = d/2 = 40 mm.
- Maximum deflection (SS + central load, at midspan): δ_max = F·L³/(48·E·I) = 5000 × 2000³/(48 × 210 000 × 2 010 619). Numerator = 5000 × 8 × 10⁹ = 4.000 × 10¹³. Denominator = 48 × 210 000 × 2 010 619 = 2.027 × 10¹³. Therefore δ_max = 4.000 × 10¹³ / 2.027 × 10¹³ = 1.97 mm.
- Peak bending moment (at midspan): M_max = F·L/4 = 5000 × 2000/4 = 2 500 000 N·mm = 2.50 kN·m.
- Maximum bending stress: σ_max = M_max·c/I = 2 500 000 × 40/2 010 619 = 100 000 000/2 010 619 = 49.7 MPa.
- Yield safety factor: SF = Sy/σ_max = 250/49.7 = 5.03. The beam is well within the elastic limit.
Illustrative example with round inputs. A solid circular bar of 80 mm diameter is heavier than a typical structural choice for a 2 m span; an I-section of similar depth achieves a much higher I and lower deflection for the same mass. Verify all results against your actual geometry, loads and the applicable material standard.
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Open the Beam DeflectionFrequently asked questions
Which standard governs beam deflection design?
Beam deflection calculations use closed-form Euler-Bernoulli beam theory, which is the analytical foundation referenced by all major structural codes. For steel structures, EN 1993-1-1 (Eurocode 3) specifies the serviceability deflection limits (typically L/250–L/500 of the span depending on use) and the strength checks (bending, shear, buckling). The deflection formulas themselves are standard analytical results; the code imposes the limits you must stay within and the load factors to apply.
Why is the cantilever deflection so much larger than simply supported for the same span?
The tip deflection of a cantilever under an end point load is F·L³/(3·E·I), while a simply supported beam under a central load deflects by F·L³/(48·E·I) — a factor of 16 difference for the same span and section. The reason is twofold: the cantilever has no support at the loaded end to resist rotation, so the full span bends; and the boundary condition at the fixed root produces a moment gradient that generates more curvature than the symmetric simply supported case. This is why cantilevers are designed with much higher section depths or much shorter spans.
Can I simply add the deflections from a point load and a UDL?
Yes — for linear elastic (Euler-Bernoulli) beams the deflections and bending moments from independent loads can be added directly by superposition. The combined midspan deflection for a simply supported beam carrying both a central point load F and a UDL w is δ_total = F·L³/(48·E·I) + 5·w·L⁴/(384·E·I). The MechanixCalc combined-loading panel does this automatically and sums the bending moments as well.
What deflection limit should I design to?
Deflection limits depend on the application. EN 1993-1-1 Table NA.1 (UK National Annex) and Annex A1.4 suggest δ ≤ L/300 to L/500 for floor beams under imposed load, L/250 for floor beams under total load, and L/200–L/250 for roof beams not supporting brittle finishes. Crane runway girders typically require δ ≤ L/600 or tighter per the crane manufacturer's requirements. Always check the limit specified by the applicable code and the client's specification.
Is the beam deflection calculator free?
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Related
- Column Buckling (Euler / EN 1993-1-1)Check slender compression members against Euler and Johnson buckling when the beam also carries axial load.
- Fatigue Analysis (Goodman / S-N)Detailed fatigue life calculation for beams and components under cyclic loading histories.
- Weld Design (EN 1993-1-8)Size the fillet or butt welds that connect the beam to its support structure.
- Column buckling — Euler and Johnson methods explained
- Fillet weld design to EN 1993-1-8